MHB Max/Min Value of f(x,y) in Range x^2+y^2≤2: Find Max/Min Value of f

[Petrus]

Calculate highest and lowest value of the function
$$f(x,y)\frac{6}{x^2+y^2+1}+3xy$$ in the range
$$\frac{1}{3}≤x^2+y^2≤2$$
Progress
$$f_x(x,y)=3y-\frac{12x}{(x^2+y^2+1)^2}$$
$$f_y(x,y)=3x-\frac{12y}{(x^2+y^2+1)^2}$$
Now when I have derivate respect to x or y I shall find critical point.
$$3y-\frac{12x}{(x^2+y^2+1)^2}=0$$
$$3x-\frac{12y}{(x^2+y^2+1)^2}=0$$

Well for the numerator:
$$3x-12y=0<=>x=4y$$
$$3y-12x=0<=>3y-48y=0 <=>y=0$$ and that means also $$x=0$$
Denominator:
$$1-(x^2+y^2+1)^2=0 <=>x=\sqrt{+-1-y^2-1^2}$$ (I will admit it did take me a while notice I had 1 minus the function :P)
well to input that x into the function is the part I am unsure too handle with.

 

[MarkFL]

Let's go back to where you have taken the first partials and equated to zero. What you want to do is combine the terms on the left by getting a common denominator, and then look at the numerators, as the denominators have no real roots. You should find 5 critical points, but one of them is not in or on the bounding region.

 

[Petrus]

$$f_y(x,y)=\frac{3x(x^2+y^2+1)-12y}{(x^2+y^2+1)}$$
Numerirator:
$$3x(x^2+y^2+1)-12y=0$$ that means $$3x-12y=0<=>x=4y$$
$$(x^2+y^2+1)-12y=0 <=> x=\sqrt{\sqrt{12y}-y^2-1}$$
Denominator:
$$(x^2+y^2+1)^2=0 <=> x=\sqrt{-y^2-1}$$ and that one we shall ignore, is this correct?

 

[MarkFL]

You should get after simplification:

$$\frac{y(x^2+y^2+1)^2-4x}{(x^2+y^2+1)^2}=0$$

$$\frac{x(x^2+y^2+1)^2-4y}{(x^2+y^2+1)^2}=0$$

We may ignore the denominators as they have no real roots, i.e., the sum of the squares of two real numbers can never be -1. So this reduces the system to:

$$y(x^2+y^2+1)^2-4x=0$$

$$x(x^2+y^2+1)^2-4y=0$$

Can we solve both equations for the same quantity, and then equate those expressions to get a relationship between $x$ and $y$?

 

[Petrus]

You should get after simplification:

$$\frac{y(x^2+y^2+1)^2-4x}{(x^2+y^2+1)^2}=0$$

$$\frac{x(x^2+y^2+1)^2-4y}{(x^2+y^2+1)^2}=0$$

We may ignore the denominators as they have no real roots, i.e., the sum of the squares of two real numbers can never be -1. So this reduces the system to:

$$y(x^2+y^2+1)^2-4x=0$$

$$x(x^2+y^2+1)^2-4y=0$$

Can we solve both equations for the same quantity, and then equate those expressions to get a relationship between $x$ and $y$?

Can you give me a tips

 

[chisigma]

Calculate highest and lowest value of the function
$$f(x,y)\frac{6}{x^2+y^2+1}+3xy$$ in the range
$$\frac{1}{3}≤x^2+y^2≤2$$
Progress
$$f_x(x,y)=3y-\frac{12x}{(x^2+y^2+1)^2}$$
$$f_y(x,y)=3x-\frac{12y}{(x^2+y^2+1)^2}$$
Now when I have derivate respect to x or y I shall find critical point.
$$3y-\frac{12x}{(x^2+y^2+1)^2}=0$$
$$3x-\frac{12y}{(x^2+y^2+1)^2}=0$$

Well for the numerator:
$$3x-12y=0<=>x=4y$$
$$3y-12x=0<=>3y-48y=0 <=>y=0$$ and that means also $$x=0$$
Denominator:
$$1-(x^2+y^2+1)^2=0 <=>x=\sqrt{+-1-y^2-1^2}$$ (I will admit it did take me a while notice I had 1 minus the function :P)
well to input that x into the function is the part I am unsure too handle with.

The problem is greatly simplified if You divide the function by 3 and the convert it in polar coordinates setting $\displaystyle x= \rho\ \cos \theta$ and $y = \rho\ \sin \theta$ so that You have to maximize/minimize the function...

$\displaystyle f(\rho, \theta)= \frac{2}{1+\rho^{2}} + \rho^{2}\ \sin \theta\ \cos \theta$ (1)

... with the condition...

$\displaystyle \frac{1}{3} \le \rho^{2} \le 2$ (2)

First we compute the partial derivatives...

$\displaystyle f_{\rho}(\rho, \theta)= \frac{4\ \rho}{(1+\rho^{2})^{2}}+ 2\ \rho\ \sin \theta\ \cos \theta$

$\displaystyle f_{\theta} (\rho,\theta)= \rho^{2}\ (\cos^{2} \theta- \sin^{2} \theta)$ (3)

... and we observe that $f_{\theta}(*,*)$ vanishes for $\displaystyle \sin \theta = \pm \cos \theta = \pm \frac{1}{\sqrt{2}}$, so that we arrive at the two equation in $\rho$...

$\displaystyle \frac{4}{(1+\rho^{2})^{2}} - 1 =0$ (4)

$\displaystyle \frac{4}{(1+\rho^{2})^{2}} + 1 =0$ (5)

The (5) has no real solutions and the only real positive solution of (4) is $\displaystyle \rho=1$, intenal os the annulus $\displaystyle \frac{1}{\sqrt{3}} \le \rho \le \sqrt{2}$ so that the point of minima or maxima are for $\displaystyle \rho=1$ and $\displaystyle \theta = \frac{\pi}{4},\ \frac{3}{4}\ \pi,\ \frac{5}{4}\ \pi,\ \frac{7}{4}\ \pi$. Further detail are left to You...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

I would solve both for $(x^2+y^2+1)^2$, and then equate them.

 

[Petrus]

We get that $$x=y$$

 

[MarkFL]

Close...we actually get $$x^2=y^2$$...

 

[Petrus]

Close...we actually get $$x^2=y^2$$...

Yeah I mean $$+-x=+-y$$
edit: progress will come soon

 

[MarkFL]

I would simply use:

$$y=\pm x$$

Now, investigate both cases, and you will find 3 critical points (in my first post in this topic, I originally said there would be 3, then edited my post to say 5, but I was initially correct although not for the right reason).

One case will return only 1 critical point, and the other will return 3, but the two cases share a point, so the total should be 3.

 

[Petrus]

I would simply use:

$$y=\pm x$$

Now, investigate both cases, and you will find 3 critical points (in my first post in this topic, I originally said there would be 3, then edited my post to say 5, but I was initially correct although not for the right reason).

One case will return only 1 critical point, and the other will return 3, but the two cases share a point, so the total should be 3.

Here is my progress $$y(y^2+y^2+1)^2+4y=0$$ so we got $$y+4y=0 <=> y=0$$ and then the hard part $$(2y^2+1)^2+4y=0 <=>4y^4+4y^2+1+4y=0$$ and if we factor out y $$y(4y^3+4y+4)+1=0 $$ and we got then $$y=-1$$ and got now problem solve that is left

 

[MarkFL]

Here is my progress $$y(y^2+y^2+1)^2+4y=0$$ so we got $$y+4y=0 <=> y=0$$...

This is incorrect...you want to factor, and then use the zero-factor property and equate both factors to zero and solve for the variable.

$$y(y^2+y^2+1)^2+4y=0$$

Notice both terms have $y$ as a factor...

What you have posted is the case $$x=-y$$, and so after this, you need to look at the case $$x=y$$.

 

[Petrus]

This is incorrect...you want to factor, and then use the zero-factor property and equate both factors to zero and solve for the variable.

$$y(y^2+y^2+1)^2+4y=0$$

Notice both terms have $y$ as a factor...

What you have posted is the case $$x=-y$$, and so after this, you need to look at the case $$x=y$$.

Ok so I get for x and y $$+-\sqrt{\frac{3}{4}}$$, $$+-\sqrt{\frac{2}{4}}$$ and $$0$$
(I got only 0 as crit point in that above and rest complex, got rest from $$x=y$$

 

[MarkFL]

I agree with all except $$x=y=\pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}$$

Can you show us the process by which you obtained these points?

 

[Petrus]

I agree with all except $$x=y=\pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}$$

Can you show us the process by which you obtained these points?

$$4y^4+4y^2=3$$ so I factour out 4y^2 and get $$4y^2(y^2+1)=3MATH] I did write wrong point... What do I do wrong?

 

[MarkFL]

Where did you get $$4y^4+4y^2=3$$ ?

edit: Okay, now I see you divided through by $y$, keeping $y=0$ in mind, and obtained:

$$4y^4+4y^2-3=0$$

So, you would factor to get:

$$(2y^2+3)(2y^2-1)=0$$

Here is the method I used to get the same roots:

In using $x=y$, you should obtain:

$$y(y^2+y^2+1)^2-4y=0$$

$$y\left((2y^2+1)^2-2^2 \right)=0$$

$$y(2y^2+3)(2y^2-1)=0$$

hence:

$$x=y=0,\,\pm\frac{1}{\sqrt{2}}$$

 

[Petrus]

Where did you get $$4y^4+4y^2=3$$ ?

edit: Okay, now I see you divided through by $y$, keeping $y=0$ in mind, and obtained:

$$4y^4+4y^2-3=0$$

So, you would factor to get:

$$(2y^2+3)(2y^2-1)=0$$

Here is the method I used to get the same roots:

In using $x=y$, you should obtain:

$$y(y^2+y^2+1)^2-4y=0$$

$$y\left((2y^2+1)^2-2^2 \right)=0$$

$$y(2y^2+3)(2y^2-1)=0$$

hence:

$$x=y=0,\,\pm\frac{1}{\sqrt{2}}$$

Now that I got my critical point how shall I work with my end point? Shall I use the trick that the one early reply use subsitate on that range so we got $$u=\frac{1}{3}$$ and $$u=2$$ and put it on orginal function and look for highest and lowest value?

 

[MarkFL]

Which of the 3 critical points is outside of the annular bounding region?

After that, you could use the suggestion made by chisigma, or you could use Lagrange multipliers with two cases, or use substitution to get a function in one variable with endpoints.

 

[Petrus]

Which of the 3 critical points is outside of the annular bounding region?

After that, you could use the suggestion made by chisigma, or you could use Lagrange multipliers with two cases, or use substitution to get a function in one variable with endpoints.

zero cause its less then 1/3

 

[MarkFL]

zero cause its less then 1/3

Correct, the origin is not on the annulus.

Now, what critical points do you find on the two boundaries?

 

[Petrus]

Can I subsitute that $$y^2=2-x^2$$ or?

 

[MarkFL]

Yes, just be mindful that you will have two cases to consider when substituting for $y$, i.e., the positive and negative roots.

 

[Petrus]

Yes, just be mindful that you will have two cases to consider when substituting for $y$, i.e., the positive and negative roots.

I get the roots $$+-\sqrt(2)$$ and one we alredy got $$+-\frac{1}{\sqrt{2}}$$ are they correct?

 

[MarkFL]

No what I mean is we have:

$$y^2=2-x^2$$ hence:

$$y=\pm\sqrt{2-x^2}$$

and so you would consider the cases:

i) $$f(x)=2+3x\sqrt{2-x^2}$$

ii) $$f(x)=2-3x\sqrt{2-x^2}$$

 

[Petrus]

No what I mean is we have:

$$y^2=2-x^2$$ hence:

$$y=\pm\sqrt{2-x^2}$$

and so you would consider the cases:

i) $$f(x)=2+3x\sqrt{2-x^2}$$

ii) $$f(x)=2-3x\sqrt{2-x^2}$$

Yeah I do that but forgot that I factour out aswell, u got à typo, it should be x not +-3x ( u factour out 3)

 

[MarkFL]

While I do make more than my share of typos it seems (Thinking), I took the original function:

$$f(x,y)=\frac{6}{x^2+y^2+1}+3xy$$

and used:

$$x^2+y^2=2$$ and $$y=\pm\sqrt{2-x^2}$$ to get:

$$f(x)=\frac{6}{2+1}\pm3x\sqrt{2-x^2}=2\pm3x\sqrt{2-x^2}$$

 

[Petrus]

Now I get
in case when $$y=\pm\sqrt{2-x^2}$$
i) $$f(x)=2+3x\sqrt{2-x^2}$$
ii) $$f(x)=2-3x\sqrt{2-x^2}$$
the roots $$x=\pm1$$ and $$x=\pm\sqrt{2}$$

in case when $$y=\pm\sqrt{\frac{1}{3}-x^2}$$
We got the roots $$x=\pm\frac{1}{\sqrt{6}}$$, $$x=\pm\frac{1}{\sqrt{3}}$$
Is this correct now...?
(Honestly I think my last step case I got wrong.)

 

[MarkFL]

I do believe you are correct. Let's look at both boundaries by using a general radius $r$ for the boundary:

$$x^2+y^2=r^2$$

and so the function becomes:

$$f(x)=\frac{6}{r^2+1}\pm3x\sqrt{r^2-x^2}$$

and so differentiating with respect to $x$, we find:

$$f'(x)=\pm3\left(x\frac{-x}{\sqrt{r^2-x^2}}+\sqrt{r^2-x^2} \right)=\pm\frac{3(r^2-2x^2)}{\sqrt{r^2-x^2}}$$

and so our critical values are:

$$x=\pm\frac{r}{\sqrt{2}},\,\pm r$$

So in the case of this problem, where the inner boundary is $$r=\frac{1}{\sqrt{3}}$$ this yields the critical values:

$$x=\pm\frac{1}{\sqrt{6}},\,\pm\frac{1}{\sqrt{3}}$$

and in the case of the outer boundary where $$r=\sqrt{2}$$ we have the critical values:

$$x=\pm1,\,\pm \sqrt{2}$$

Good work, Petrus! (Yes)

 

[Petrus]

I do believe you are correct. Let's look at both boundaries by using a general radius $r$ for the boundary:

$$x^2+y^2=r^2$$

and so the function becomes:

$$f(x)=\frac{6}{r^2+1}\pm3x\sqrt{r^2-x^2}$$

and so differentiating with respect to $x$, we find:

$$f'(x)=\pm3\left(x\frac{-x}{\sqrt{r^2-x^2}}+\sqrt{r^2-x^2} \right)=\pm\frac{3(r^2-2x^2)}{\sqrt{r^2-x^2}}$$

and so our critical values are:

$$x=\pm\frac{r}{\sqrt{2}},\,\pm r$$

So in the case of this problem, where the inner boundary is $$r=\frac{1}{\sqrt{3}}$$ this yields the critical values:

$$x=\pm\frac{1}{\sqrt{6}},\,\pm\frac{1}{\sqrt{3}}$$

and in the case of the outer boundary where $$r=\sqrt{2}$$ we have the critical values:

$$x=\pm1,\,\pm \sqrt{2}$$

Good work, Petrus! (Yes)

Honestly you are a really great explainer and person. It did take like 20+ post and you keep helping while you don't have to, I have always admired you for so long time. I wish their was more people like you in this world. I have had one really good math teacher that sometimes did spend his free time to learn math to other student that had problem. He did also become great friend with evryone and then their is teacher that don't bother if someone don't understand. Well it's not their problem if someone don't understand but it shows that their is people with big heart and that is something that admired. Unfortently you can't show how greatful you are in internet, but thanks a lot Mark:)
edit: I forgot to thank chisigma aswell, sorry.